Chapter 15: Problem 26

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$ y^{\prime \prime}+2 y^{\prime}+10 y=-6 e^{-t} \sin 3 t, \quad y_{0}=0, \quad y_{0}^{\prime}=1 $$

Short Answer

Expert verified

The solution is $ y(t) = -17e^{-t} \sin(3t) $.

Step by step solution

Take the Laplace Transform of both sides

Take the Laplace transform of both sides of the differential equation y'' + 2y' + 10y = -6e^{-t} \sin(3t)Using the linearity property of the Laplace transform and known transforms, we get:\[ \mathcal{L} \{ y''+ 2y' + 10y \} = \mathcal{L} \{ -6e^{-t} \sin(3t) \} \]

Use Laplace transform properties on the left-hand side

Apply the Laplace transform properties for derivatives: \[ \mathcal{L} \{ y'' \} = s^2Y(s) - sy(0) - y'(0) \ \mathcal{L} \{ y' \} = sY(s) - y(0) \ \mathcal{L} \{ y \} = Y(s) \]Since y(0) = 0 and y'(0) = 1, substitute these values:\[ s^2Y(s) - 0 - 1 + 2(sY(s) - 0) + 10Y(s) \]This simplifies to:\[ s^2Y(s) - 1 + 2sY(s) + 10Y(s) \]

Combine terms on the left-hand side

Combine all the terms involving $Y(s)$ on the left side:\[ (s^2 + 2s + 10)Y(s) - 1 \]

Use Laplace transform for the right-hand side

Use the known Laplace transform for the product of functions involving an exponential and a sine function, $ \mathcal{L} \{ e^{-at} \sin(bt) \} = \frac{b}{(s+a)^2 + b^2} $ In our case, with a = 1 and b = 3:\[ \mathcal{L} \{ -6e^{-t} \sin(3t) \} = -6 \cdot \frac{3}{(s+1)^2 + 9} = -6 \cdot \frac{3}{s^2 + 2s + 10} \]This simplifies to:\[ \mathcal{L} \{ -6e^{-t} \sin(3t) \} = -\frac{18}{s^2 + 2s + 10} \]

Set up the equation in terms of Y(s)

Combine the left and right side transforms to get:\[ (s^2 + 2s + 10)Y(s) - 1 = -\frac{18}{s^2 + 2s + 10} \]

Solve for Y(s)

Rearrange the equation to solve for $Y(s)$:\[ (s^2 + 2s + 10)Y(s) = -\frac{18}{s^2 + 2s + 10} + 1 \]Multiply both sides by $(s^2 + 2s + 10)$ to isolate $Y(s)$:\[ Y(s) = -18/(s^2 + 2s + 10) + \frac{1}{(s^2 + 2s + 10)} \]Combine to get:\[ Y(s) = \frac{-18 + 1}{s^2 + 2s + 10} = \frac{-17}{s^2 + 2s + 10} \]

Take the inverse Laplace transform

Apply the inverse Laplace transform to find $y(t)$:\[ y(t) = \mathcal{L}^{-1} \left\{ \frac{-17}{s^2 + 2s + 10} \right\} \]Using the known inverse transforms, recognize this as a second order polynomial in the denominator, which corresponds to a damped sine/cosine function. Thus, the solution is:\[ y(t) = -17e^{-t} \sin(3t) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laplace Transform Properties

Understanding the properties of Laplace Transforms is crucial for solving differential equations like the one we tackled. The main properties used include: linearity, which allows the transform of a sum to be the sum of the transforms; and the transformation of derivatives, which simplifies the problem by turning it into an algebraic equation. We use:

\text{The Laplace transform of a function:} \ \ $ \mathcal{L} \{ f(t) \} = \int_0^\infty f(t)e^{-st} dt $
\text{First Derivative:}\ \ $ \mathcal{L} \{ f'(t) \} = sF(s) - f(0) $
\text{Second Derivative:}\ \ $ \mathcal{L} \{ f''(t) \} = s^2F(s) - sf(0) - f'(0) $

Inverse Laplace Transform

After solving for Y(s) in the Laplace domain, we need to revert it back to the time domain using the inverse Laplace transform. This step is essential to get the final solution to the differential equation. Using the properties of inverse transforms:To find $ y(t) $ for simple fractions, like $ \frac{-17}{s^2 + 2s + 10} $, we need to recognize patterns in our tables of transforms. The expression $ \frac{1}{s^2 + 2s + 10} $ typically converts to a damped sine or cosine function in time domain.

Second Order Polynomial

In the solution process, dealing with second order polynomials in the denominator of Laplace transforms is common. A second order polynomial looks like: \ (s^2 + bs + c)\Recognizable patterns for the inverse transform are crucial. For instance, given a:

$ \frac{1}{(s+\text{constant})^2 + (\text{positive constant})^2} \Rightarrow e^{-at} \sin(bt) $

Damped Sine Function

The term \text{damped sine function} appears due to the form of the inverse transform. It involves a sine function multiplied by an exponential factor, showing a decay over time. For instance:

$ y(t) = e^{-at}\text{sin(bt)} $

In our problem, the Laplace transform results in $ e^{-t} \sin(3t) $, which is a damped sine wave with a decay constant of 1 and angular frequency of 3. These concepts help provide both the magnitude and oscillatory nature of solutions to differential equations.